Bài 1 (Trang 63 SGK Toán 7, Bộ Cánh diều, Tập 2)

a) $\frac{1}{2}{x}^2 . \frac{6}{5}{x}^3;$

b) $y^2\left(\frac{5}{7}{y}^3 - 2y^2 + 0,25\right);$

c) (2x² + x + 4)(x² - x - 1);

d) (3x - 4)(2x + 1) - (x - 2)(6x + 3).

Hướng dẫn giải

a) $\frac{1}{2}{x}^2 . \frac{6}{5}{x}^3 =\frac{1}{2} . \frac{6}{5} . x^2 . x^3 = \frac{3}{5}{x}^5$

b) $y^2\left(\frac{5}{7}{y}^3 - 2y^2 + 0,25\right)$

$= y^2 . \frac{5}{7}{y}^3 - y^2 . 2y^2 + y^2 . 0,25$

$= \frac{5}{7}{y}^5 - 2y^4 + 0,25y^2$

c) (2x² + x + 4)(x² - x - 1)

= 2x² . x² - 2x² . x - 2x² . 1 + x . x² - x . x - x . 1 + 4 . x² - 4 . x - 4 . 1

= 2x⁴ - 2x³ - 2x² + x³ - x² - x + 4x² - 4x - 4

= 2x⁴ + (-2x³ + x³) + (-2x² - x²+ 4x²) + (-x - 4x) - 4

= 2x⁴ - x³ + x² - 5x - 4.

d) (3x - 4)(2x + 1) - (x - 2)(6x + 3)

= 3x . 2x + 3x . 1 - 4 . 2x - 4 . 1 - (x . 6x + x . 3 - 2 . 6x - 2 . 3)

= 6x² + 3x - 8x - 4 - (6x² + 3x - 12x - 6)

= 6x² - 5x - 4 - (6x² - 9x - 6)

= 6x² - 5x - 4 - 6x² + 9x + 6

= (6x² - 6x²) + (-5x + 9x) + (-4 + 6)

= 4x + 2.